All categories

3 years ago

19-13p=-17p-5 check solution

1 answer:

5 0

$19-13p=-17p-5\ \ \ \ |add\ 13p\ to\ each \ side\\\\
19=-17p+13p-5\\\\
19=-4p-5\ \ \ \ |add\ 5\\\\
24=-4p\ \ \ \ |divide\ by\ -4\\\\p=-6\\\\Solution\ is\ p=-6.$

You might be interested in

Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters

Mumz [18]

The various answers to the question are:

To answer 90% of calls instantly, the organization needs four extension lines.
The average number of extension lines that will be busy is Four
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$$

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$$

Generally, the equation for is mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

The probability that a call will receive a busy signal if four extensions lines are used is,

$P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$$

Therefore, the average number of extension lines that will be busy is Four

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

Read more about extension lines

brainly.com/question/13362603

#SPJ1

8 0

2 years ago

An employee at a company that assembles chandeliers is packing boxes for shipping. In the first box, he packed 5 small chandelie

valkas [14]

Answer:

Step-by-step explanation:

5 0

3 years ago

Can someone explain how to solve this and the solution? Please and thank you :)

Verizon [17]

25 > 2x + 8 - 66

first simplify all numbers

25 > 2x + (8 - 66)
25 > 2x - 58

Then add 58 to both sides

25 (+58) > 2x - 58 (+58)
83 > 2x

Finally divide 2 from both sides to isolate the x

83/2 > 2x/2

41.5 > x

hope this helps

3 0

3 years ago

A baby weights 18 pounds at her four month appointment. Six months later she weights 24 pounds. By what percentage did the baby’

Anna11 [10]

Answer: 25%

Step-by-step explanation:

Given : Previous baby's weight = 18 pounds

New baby's weight = 24 pounds.

Increase in weight = New weight -Previews weight

=24 pounds - 18 pounds =6 pounds

Percentage increase in baby's weight = $=\dfrac{\text{Increase in baby's weight}}{\text{previous weight}}\times100$

$\\\\=\dfrac{6}{24}\times100=25\%$

Hence, the percentage increase in baby's weight = 25%

4 0

3 years ago

Ermias and Jeremiah both leave the bookstore at the same time, but in opposite directions. If Jeremiah travels 8 mph faster than

leva [86]

Using proportions, it is found that Ermias is traveling at a rate of 11 mph and Jeremiah at a rate of 19 mph.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

Jeremiah travels 8 mph faster than Ermias, hence their velocities are:

x, x + 8

They travel in opposite directions, hence in one hour they are 2x + 8 apart.

In eight hours, the distance is 8(2x + 8), which is of 240 miles, hence:

8(2x +8) = 240

2x + 8 = 30

2x = 22

x = 11.

Then:

x = 11 -> Ermias.

x = 11 + 8 = 19 -> Jeremiah.

More can be learned about proportions at brainly.com/question/24372153

#SPJ1

3 0

2 years ago